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Homework 2
Chapter 19
Problem 31. A d = 40.0 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The
flux in this position is measured to be ΦE = 5.20 · 105 N·m2 /C. What is the magnitude of the electric field?
ΦE = EA
E=
(1)
ΦE
ΦE
=
=
A
π(d/2)2
5.20 · 105
N·m2 /C
π · (0.200 m)2
= 4.14 · 106 N/C
(2)
Problem 36. An m = 10.0 g piece of Styrofoam carries a net charge of q = −0.700 µC and floats above the center of a large horizontal
sheet of plastic that has a uniform charge density σ on it’s surface. Find σ.
Because the Styrofoam is floating in equilibrium, the sum of forces in the vertical direction must be zero. So
σ
2ε0
2ε0 mg 2 · 8.54 · 10−12 C2 /N·m2 · 0.0100 kg · 9.80 m/s2
σ=
=
= 2.39 · 10−6 C/m2
q
−0.700 · 10−6 C
Fg = mg = FE = qE = q
(3)
(4)
Problem 55. Four identical point charges (q = +10.0 µC) are located on the corners of a rectangle as shown in Figure P19.55. The
dimensions of the rectangle are L = 60.0 cm and W = 15.0 cm. Calculate the magnitude and direction of the resultant electric force
exerted on the charge at the lower left corner by the other three charges.
This is just a jazzed up version of Problem 15 from recitation. The unit vector r̂ diagonally across from the upper right is given by
r̂ = cos θî + sin θĵ
θ = arctanW /L + 180◦ = 194◦
cos θ = −0.970
sin θ = −0.243
(5)
(6)
(7)
(8)
So the electric field in the lower left corner is given by
q
q
qi
q
E = ke ∑ 2 r̂i = ke
(−
î)
+
(cos
θ
î
+
sin
θ
ĵ)
+
(−
ĵ)
L2
(L2 +W 2 )
W2
i ri
cos θ
1
sin θ
1
−
î +
−
ĵ
= −ke q
L2 L2 +W 2
W 2 L2 +W 2
(9)
(10)
So the magnitude of E is given by
s
E = ke q
L−2 −
cos θ
L2 +W 2
2
+ W −2 −
sin θ
L2 +W 2
2
= 4.08 · 106 N/C
(Remembering to convert L and W to meters.) And the direction θ (measured counter clockwise from î) of E is given by
!
θ
−W −2 + L2sin
+W 2
+ 180◦ = 263◦
θ = arctan
θ
−L−2 + L2cos
+W 2
(11)
(12)
Where the +180◦ is because the tangent has a period of 180◦ , and the angle we want is in the backside 180◦ .
F = qE so the direction of F is the same as the direction of E. The magnitude of F is given by
F = 10.0 · 10−6 C · 4.08 · 106 N/C = 40.8 N
1
(13)